Let S = \ x ∈ R: 0 < x < 1and2 ^ - 1 ((1 - x)/(1 + x)) = ^ - 1 ( 1...

Given, $0 < x < 1$

$2\tan^{- 1}\left(\dfrac{1 - x}{1 + x}\right) = \cos^{- 1}\left(\dfrac{1 - {x}^{2}}{1 + {x}^{2}}\right)$

Put $x = \tan \left(\theta\right); 0 < \theta < \dfrac{\pi}{4}$

$2\tan^{- 1}\left(\dfrac{1 - tan\theta}{1 + tan\theta}\right) = \cos^{- 1}\left(\dfrac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta}\right)$

$⇒ 2\tan^{- 1}\left(\tan \left(\dfrac{\pi}{4} - \theta\right)\right) = \cos^{- 1} ( \cos 2\theta )$

$2\left(\dfrac{\pi}{4} - \theta\right) = 2\theta ⇒ \theta = \dfrac{\pi}{8}$

Put $\theta = \tan^{- 1}x$ we get,

$x = tan\dfrac{\pi}{8} ∴ x = \sqrt{2} - 1 ≃ 0 . 414 < \dfrac{1}{2}$